Matrix and algorithm "spiral"

Question

i wanted ask if there some algorithm ready, that allowed me to do this: i have a matrix m (col) x n (row) with m x n elements. I want give position to this element starting from center and rotating as a spiral, for example, for a matrix 3x3 i have 9 elements so defined:

 5  6  7
 4  9  8
 3  2  1 

or for una matrix 4 x 3 i have 12 elements, do defined:

  8   9  10  1
  7  12  11  2
  6   5   4  3

or again, a matrix 5x2 i have 10 elements so defined:

    3  4
    7  8
   10  9
    6  5 
    2  1

etc. I have solved basically defining a array of integer of m x n elements and loading manually the value, but in generel to me like that matrix maked from algorithm automatically. Thanks to who can help me to find something so, thanks very much.

UPDATE

This code, do exactely about i want have, but not is in delphi; just only i need that start from 1 and not from 0. Important for me is that it is valid for any matrics m x n. Who help me to translate it in delphi?

(defun spiral (rows columns)  
  (do ((N (* rows columns))       
    (spiral (make-array (list rows columns) :initial-element nil))      
    (dx 1) (dy 0) (x 0) (y 0)    
   (i 0 (1+ i)))     
 ((= i N) spiral)   
 (setf (aref spiral y x) i)
    (let ((nx (+ x dx)) (ny (+ y dy)))  
    (cond       
       ((and (< -1 nx columns)       
       (< -1 ny rows)           
       (null (aref spiral ny nx)))    
    (setf x nx 
          y ny))  
     (t (psetf dx (- dy)                 
       dy dx)       
    (setf x (+ x dx)             
     y (+ y dy))))))) 


> (pprint (spiral 6 6))

#2A ((0  1  2  3  4  5)
    (19 20 21 22 23  6)
    (18 31 32 33 24  7)
    (17 30 35 34 25  8)
    (16 29 28 27 26  9)
    (15 14 13 12 11 10))


> (pprint (spiral 5 3))

#2A  ((0  1 2)
     (11 12 3)
     (10 13 4)
      (9 14 5)
      (8 7 6))

Thanks again very much.

Answer 1

Based on the classic spiral algorithm. supporting non-square matrix:

program SpiralMatrix;
{$APPTYPE CONSOLE}
uses
  SysUtils;

type
  TMatrix = array of array of Integer;

procedure PrintMatrix(const a: TMatrix);
var
  i, j: Integer;
begin
  for i := 0 to Length(a) - 1 do
  begin
    for j := 0 to Length(a[0]) - 1 do
      Write(Format('%3d', [a[i, j]]));
    Writeln;
  end;
end;

var
  spiral: TMatrix;
  i, m, n: Integer;
  row, col, dx, dy,
  dirChanges, visits, temp: Integer;
begin
  m := 3; // columns
  n := 3; // rows
  SetLength(spiral, n, m);
  row := 0;
  col := 0;
  dx := 1;
  dy := 0;
  dirChanges := 0;
  visits := m;
  for i := 0 to n * m - 1 do
  begin
    spiral[row, col] := i + 1;
    Dec(visits);
    if visits = 0 then
    begin
      visits := m * (dirChanges mod 2) + n * ((dirChanges + 1) mod 2) - (dirChanges div 2) - 1;
      temp := dx;
      dx := -dy;
      dy := temp;
      Inc(dirChanges);
    end;
    Inc(col, dx);
    Inc(row, dy);
  end;
  PrintMatrix(spiral);
  Readln;
end.

3 x 3:

1  2  3
8  9  4
7  6  5

4 x 3:

 1  2  3  4
10 11 12  5
 9  8  7  6

2 x 5:

 1  2
10  3
 9  4
 8  5
 7  6

Answer 2

There you go!!! After 30some syntax errors...

On ideone.com, I ran it with some tests and it seems to work fine. I think you can see the output there still and run it yourself...

I put some comments in the code. Enough to understand most of it. The main navigation system is a little bit harder to explain. Briefly, doing a spiral is going in first direction 1 time, second 1 time, third 2 times, fourth 2 times, fifth 3 times, 3, 4, 4, 5, 5, and so on. I use what I called a seed and step to get this behavior.

program test;

var
    w, h, m, n, v, d : integer; // Matrix size, then position, then value and direction.
    spiral : array of array of integer; // Matrix/spiral itself.
    seed, step : integer; // Used to travel the spiral.

begin
    readln(h);
    readln(w);
    setlength(spiral, h, w);
    v := w * h; // Value to put in spiral.
    m := trunc((h - 1) / 2);  // Finding center.
    n := trunc((w - 1) / 2);
    d := 0; // First direction is right.

    seed := 2;
    step := 1;

    // Travel the spiral.
    repeat
        // If in the sub-spiral, store value.
        if ((m >= 0) and (n >= 0) and (m < h) and (n < w)) then
        begin
            spiral[m, n] := v;
            v := v - 1;
        end;

        // Move!
        case d of
            0: n := n + 1;
            1: m := m - 1;
            2: n := n - 1;
            3: m := m + 1;
        end;

        // Plan trajectory.
        step := step - 1;
        if step = 0 then
        begin
            d := (d + 1) mod 4;
            seed := seed + 1;
            step := trunc(seed / 2);
        end;
    until v = 0;

    // Print the spiral.
    for m := 0 to (h - 1) do
    begin
        for n := 0 to (w - 1) do
        begin
            write(spiral[m, n], ' ');
        end;
        writeln();
    end;

end.

If you really need that to print text spirals I'll let you align the numbers. Just pad them with spaces.

EDIT:

Was forgetting... In order to make it work on ideone, I put the parameters on 2 lines as input. m, then n.

For example:

5
2

yields

3 4 
7 8 
10 9 
6 5 
2 1 

Answer 3

Even though the question is already answered, this is an alternative solution (arguably simpler). The solution is in python (using numpy for bidimendional arrays), but can be easily ported.

The basic idea is to use the fact that the number of steps is known (m*n) as end condition, and to properly compute the next element of the loop at each iteration:

import numpy as np

def spiral(m, n):
    """Return a spiral numpy array of int with shape (m, n)."""
    a = np.empty((m, n), int)
    i, i0, i1 = 0, 0, m - 1
    j, j0, j1 = 0, 0, n - 1
    for k in range(m * n):
        a[i, j] = k
        if   i == i0 and     j0 <= j <  j1: j += 1
        elif j == j1 and     i0 <= i <  i1: i += 1
        elif i == i1 and     j0 <  j <= j1: j -= 1
        elif j == j0 and 1 + i0 <  i <= i1: i -= 1
        else:
            i0 += 1
            i1 -= 1
            j0 += 1
            j1 -= 1
            i, j = i0, j0
    return a

And here some outputs:

>>> spiral(3,3)
array([[0, 1, 2],
       [7, 8, 3],
       [6, 5, 4]])
>>> spiral(4,4)
array([[ 0,  1,  2,  3],
       [11, 12, 13,  4],
       [10, 15, 14,  5],
       [ 9,  8,  7,  6]])
>>> spiral(5,4)
array([[ 0,  1,  2,  3],
       [13, 14, 15,  4],
       [12, 19, 16,  5],
       [11, 18, 17,  6],
       [10,  9,  8,  7]])
>>> spiral(2,5)
array([[0, 1, 2, 3, 4],
       [9, 8, 7, 6, 5]])

Answer 4

Here's the commented JavaScript implementation for what you're trying to accomplish.

// return an array representing a matrix of size MxN COLxROW
function spiralMatrix(M, N) {
var result = new Array(M * N);
var counter = M * N;
// start position
var curCol = Math.floor((M - 1) / 2);
var curRow = Math.floor(N / 2);
// set the center
result[(curRow * M) + curCol] = counter--;
// your possible moves RIGHT, UP, LEFT, DOWN * y axis is flipped
var allMoves = [[1,0], [0,-1], [-1,0], [0,1]];
var curMove = 0;
var moves = 1; // how many times to make current Move, 1,1,2,2,3,3,4,4 etc
// spiral
while(true) {
 for(var i = 0; i < moves; i++) {
  // move in a spiral outward counter clock-wise direction
  curCol += allMoves[curMove][0];
  curRow += allMoves[curMove][1];
  // naively skips locations that are outside of the matrix bounds
  if(curCol >= 0 && curCol < M && curRow >= 0 && curRow < N) {
   // set the value and decrement the counter
   result[(curRow * M) + curCol] = counter--;
   // if we reached the end return the result
   if(counter == 0) return result;
  }
 }
 // increment the number of times to move if necessary UP->LEFT and DOWN->RIGHT
 if(curMove == 1 || curMove == 3) moves++;
 // go to the next move in a circular array fashion
 curMove = (curMove + 1) % allMoves.length;
}
}

The code isn't the most efficient, because it walks the spiral naively without first checking if the location it's walking on is valid. It only checks the validity of the current location right before it tries to set the value on it.